Regarding the chosen answer for: How does this OP-AMP non-inverting amplifier work?
The following circuit is given, and it says that $$F_c = \frac{1}{2\pi R_1C_1}$$
simulate this circuit – Schematic created using CircuitLab
However, this doesn't seem right to me, even though I have found the same example and formula elsewhere on the web (e.g., "Non-inverting amplifier – alternative" at http://stompville.co.uk/?p=470).
To me it doesn't take into account the value of the op amp feedback resistor (\$R_2\$ in the sketch above).
Would the cut-off frequency not be based upon (\$R_1+R_2\$)? If not, why not?
My point is that the only way the capacitor can charge or discharge is through the output of the op-amp, so surely the feedback resistor is also critical?
Best Answer
The cutoff frequency is the frequency at which the gain has fallen by -3dB over the gain at high frequencies.
(for the purposes of the below discussion, we'll ignore the high pass network R and C since the effects of R1/C1 are being examined)
If the signal was applied to C1 (rather than C1 being grounded) R2 would not matter at all- gain would be -R2/R1 for high frequencies. When the reactance of the C1 equals the resistance of R1 in magnitude you have a reduction in voltage of 1/sqrt(2) which is -3dB. R2 will determine the gain, but not the cutoff frequency.
However it's a bit more complex in this case, because the op-amp is being fed from the non-inverting input, and the gain is actually 1 + R2/R1 at high frequencies, so the frequency at which the gain is reduced by 1/sqrt(2) will actually depend on R2.
In an extreme case, if you short R2, it becomes a voltage follower and then Fc is limited only by the amplifier!